Find the minimum and maximum values of the function f(x)= 2/3 sin pix on the interval [1,2] (domain).
I got (1.5, -2/3) as the minimum and no maximum on the interval. Is it correct? Thank you.
correct.
I bet you mean
f(x) = (2/3) sin(pi x)
f'(x) = (2/3) pi cos(pi x)
f'(x) = 0 for max or min
that is when cos (pi x) = 0
x = 1/2 and x = 3/2
if x = 1/2
y = (2/3)sin (pi/2) = 2/3
if x = 3/2
y = (2/3) sin (3 pi/2) = -2/3
PS
You can be sure that the max and min of a sin function are +1 and -1
sorry, x = 1/2 not in domain so you are right
Thank you both! I appreciate it. :)
To find the minimum and maximum values of the function f(x) = (2/3)sin(πx) on the interval [1, 2], you first need to find the critical points of the function within this interval. Critical points occur where the derivative of the function is either zero or undefined.
Let's start by finding the derivative of f(x):
f'(x) = (2/3) * d/dx (sin(πx)).
Applying the chain rule, we have:
f'(x) = (2/3) * π * cos(πx).
Next, let's solve for f'(x) = 0 within the interval [1, 2]:
(2/3) * π * cos(πx) = 0.
Since π ≠ 0, we can divide both sides by π/3 to get:
cos(πx) = 0.
The critical points occur when cos(πx) = 0. In other words, x must be an odd multiple of 1/2, because cos(π/2) = 0, cos(3π/2) = 0, cos(5π/2) = 0, and so on.
Within the interval [1, 2], the only critical point is at x = 3/2.
Now, let's evaluate the function at the endpoints and the critical point:
f(1) = (2/3)sin(π) = 0.
f(2) = (2/3)sin(2π) = 0.
f(3/2) = (2/3)sin(π(3/2)) = (2/3)sin(3π/2) = -2/3.
Therefore, the minimum value of f(x) on the interval [1, 2] is -2/3, occurring at x = 3/2. However, there is no maximum value since the function remains at 0 at both endpoints within the interval.
So, your solution (1.5, -2/3) for the minimum is correct, but there is no maximum on the interval.